= Exponential-gap slider bearing
{title2=$h=e^{-\alpha x},\quad x_m=-\alpha^{-1}\log\frac{3(e^{2\alpha}-1)}{2(e^{3\alpha}-1)}$}
In a dimensionless thin bearing with lower-wall speed one, stationary upper wall and gap $h(x)$, <lubrication theory> gives $u=1-Y/h+(p'/2)Y(Y-h)$ and constant volume flux $q=h/2-h^3p'/12$. Equal endpoint pressures, chosen as zero at $x=0,1$, imply $(h^3p'-6h)'=0$.
For a converging exponential gap $h=e^{-\alpha x}$, $\alpha>0$, integration gives $p'=6e^{3\alpha x}(e^{-\alpha x}-M)$ with $M=3(e^{2\alpha}-1)/(2(e^{3\alpha}-1))$. This constant is the positive weighted average $\int h h^{-3}\,dx/\int h^{-3}\,dx$, so $e^{-\alpha}<M<1$. It follows that the pressure first rises, reaches a unique maximum at the displayed $x_m$, and falls back to zero. The interior pressure is positive and supports a load. Expansion gives $x_m=1/2+5\alpha/24+O(\alpha^3)$ as $\alpha\to0$.
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